Question

A statistician selected a sample of 16 accounts receivable and determined the sample standard deviation of $400. she reported that the sample information indicated the mean of the population ranges from $4739.80 to $5260.20. she did NOT report what confidence coefficient she had used. determine the confidence coefficient that was used.

a. find the mean of the confidence interval

b. find the margin of error

c. using the information find the confidence coefficient (level)

Answer 1

Statistician selected a sample of 16 accounts and the sample standard deviation is $400.

If the mean ranges from $4739.80 and $5260.20. Then we have cofidence interval as (4739.80, 5260.20).

Let us consider null and alternative hypothesis as

The confidence interval is given by

Therefore we have 4739.80 and 5260.20  as the lower and upper limit of the interval and the difference of its mean is the margin of error which is 260.2. Thus

a) Mean of the confidence interval is 5000.

b) The margin of error is 260.2

c) The critical value is obtsined by comparing the t value for given degree of frredom i.e. 15, therefore we have

level of significance =0.1=1%. and the confidence level is (100-1)%=99%.